Markus  6,050
Recognized Expert Expert
Is that normal behaviour? That is, x is of type char and when trying to then assign a number to x, no type errors are raised? However, if it was the other way around, char into int, it wouldn't compile.
Does this have to do with the memory sizes of types?
Excuse my ignorance, I'm a little old fool who's only ever really worked with dynamic 'languages'. :D
4 2600  JosAH 11,448
Recognized Expert MVP @Markus
From the Standard:
When an integer is demoted to an unsigned integer with smaller
size, the result is the nonnegative remainder on division by the
number one greater than the largest unsigned number that can be
represented in the type with smaller size. When an integer is demoted
to a signed integer with smaller size, or an unsigned integer is
converted to its corresponding signed integer, if the value cannot be
represented the result is implementation-defined.
No warning is issued so you can do what you did in an implementation defined way.
kind regards,
Jos
ps. Old?
Markus 6,050
Recognized Expert Expert @JosAH
Young wouldn't have worked...
But I am little.
Oh, and thanks.
Mark.
donbock 2,426
Recognized Expert Top Contributor @Markus
I would have expected that to work just fine. What compiler message did you get?
Markus is talking about automatic integer promotions:
If the value can be put in an int, then it is converted to an int
otherwise, it is converted to an unsigned int.
Now, if either operand is unsigned long, then the other is converted to unsigned long.
Now, if either operand is long, then the other is converted to long.
Now, if either operand is an unsigned int, then the other is converted to unsigned int.
Otherwise, both are ints.
Then the operator is performed and the type of the result is the type of the operands.
So, that char is converted to int. The calculation is done as an int and the result put in a char. It is your responsibility to insure the char can hold the result.
Likewise, there are separate automatic promotions for floating point.
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